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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 11 101 Copyright © Orchard Publications Solutions to End-of-Chapter Exercises The plot shown below is the phase (not magnitude) response. 5 . We will use MATLAB for all computations. % MFB 2nd order Bessel filter, T0=100 microseconds, K=2 % PART I Find resistor values T0=100*10^( 6); K=2; C1=10^( 8); C2=2*10^( 9); a=3; b=3; w0=12/(13*T0);. .. R2=(2*(K+1))/((a*C1+sqrt(a^2*C1^2 4*b*C1*C2*(K+1)))*w0); R1=R2/K;. .. R3=1/(b*C1*C2*R2*w0^2); fprintf(' \n'); fprintf('R1 = %5.0f Ohms \t',R1);. .. fprintf('R2 = %5.0f Ohms \t',R2); fprintf('R3 = %5.0f Ohms \t',R3) R1 = 7486 Ohms R2 = 14971 Ohms R3 = 13065 Ohms We choose standard resistors as close as possible to those found above. These are shown in the MATLAB script below. Part II of the script is as follows: % % PART II Plot with standard resistors R1=7.5 K, R2=15.0 K, R3=13.0 K % K=2; a=3; b=3; C1=10^( 8); C2=2*10^( 9); f=1:10:100000; w=2*pi*f; R1=7500;. .. R2=15000; R3=13000; w0=(2*(K+1))/((a*C1+sqrt(a^2*C1^2 4*b*C1*C2*(K+1)))*R2);. .. s=w*j; Gw=(3.*K.*w0.^2)./(s.^2+a.*w0.*s+b.*w0.^2);. .. semilogx(f,angle(Gw).*180./pi); xlabel('Frequency, Hz');. .. ylabel('Phase Angle in degrees'); title('2nd Order Bessel Filter Response'); grid The plot shown below is the phase (not magnitude) response. This filter has very good phase response but poor magnitude response. The group delay (the slope at a particular frequency) is practically flat at frequencies near DC. 10 1 10 2 10 3 10 4 10 5 -200 -150 -100 -50 0 50 100 150 200 Frequency, Hz - log scale Phase Angle in degrees 2nd Order All-Pass Filter Phase Response

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Chapter 11 Analog and Digital Filters 11 102 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications 6 . From (11.24), Page 11 14, and with and , we obtain Then, We can use DeMoivre’s theorem to find the roots of but we will use MATLAB instead. syms s; y=solve('s^8+1=0'); fprintf(' \n'); disp('s1 = '); disp(simple(y(1)));. .. disp('s2 = '); disp(simple(y(2))); disp('s3 = '); disp(simple(y(3)));. .. disp('s4 = '); disp(simple(y(4))); disp('s5 = '); disp(simple(y(5)));. .. disp('s6 = '); disp(simple(y(6))); disp('s7 = '); disp(simple(y(7)));. .. disp('s8 = '); disp(simple(y(8))) s1 = 1/2*2^(3/4)*(1+i)^(1/2) s2 = -1/2*2^(3/4)*(1+i)^(1/2) s3 = 1/2*i*2^(3/4)*(1+i)^(1/2) s4 = -1/2*i*2^(3/4)*(1+i)^(1/2) s5 = 1/2*i*2^(3/4)*(-1+i)^(1/2) 10 0 10 1 10 2 10 3 10 4 10 5 -180 -160 -140 -120 -100 -80 -60 -40 -20 0 Frequency, Hz Phase Angle in degrees 2nd Order Bessel Filter Response A 2 ω () 1 ωω C 2k 1 + ----------------------------------- = ω C 1 rad s = k4 = A 2 ω 1 ω 8 1 + --------------- = Gs 1 s 8 1 + ------------- = s 8 1 + s 1 12 2 3 4 1j + ⋅⋅ = s 2 2 3 4 + = s 3 j2 3 4 + = s 4 j 2 3 4 + = s 5 3 4 1 j + =
Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 11 103 Copyright © Orchard Publications Solutions to End-of-Chapter Exercises s6 = -1/2*i*2^(3/4)*(-1+i)^(1/2) s7 = 1/2*2^(3/4)*(-1+i)^(1/2) s8 = -1/2*2^(3/4)*(-1+i)^(1/2) Since we are only interested in the poles of the left half of the plane, we choose the roots , , , and . To express the denominator in polynomial form we use the following MATLAB script: denGs=(s s2)*(s s4)*(s s6)*(s s8); r=vpa(denGs,4) r = (s+.9240+.3827*i)*(s+.3827 .9240*i)*(s+.9240 .3827*i)*(s+.3827+.9240*i) expand(r) ans = s^4+2.6134*s^3+3.41492978*s^2+2.614014906886*s +1.0004706353613841 and thus 7 . From (11.49), Page 11 26, (1) and with and , we find from (11.48), Page 11 26, that Also, from (11.56), Page 11

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